proof: derivative for (ax + b)^n

[red and white kop!]

So the title says it all:
Write down the expansion of (ax + b)^n where n is a positive integer. Differentiate the result with respect to x. Show that the derivative is na((ax + b)^(n-1))

So I'm not even sure what I'm looking for here, or how much help I'll get here. This obviously isn't a test question or a question that is going to be corrected, and what would actually help me most is a proof or link to a proof so that I can see where I'm wrong. I'm not even sure I'm wrong. I've just come to a point where its hard to wrap my head around whats going on.
So I've got to the point where I have expanded (ax + b)^n, I have differentiated it and I have (a) as a factor of (ax + b)^(n-1), except with (n) as a factor of the first term (ax)^(n-1). Now there's no way I can post all of my work, which includes Pascal notation or whatever the conventional term is.
Anyway I'd be thankful if someone having the appropriate software could provide me with a proof so I can see where I've gone wrong.
Thanks

 

[SAMUELK]

So the title says it all:
Write down the expansion of (ax + b)^n where n is a positive integer. Differentiate the result with respect to x. Show that the derivative is na((ax + b)^(n-1))

So I'm not even sure what I'm looking for here, or how much help I'll get here. This obviously isn't a test question or a question that is going to be corrected, and what would actually help me most is a proof or link to a proof so that I can see where I'm wrong. I'm not even sure I'm wrong. I've just come to a point where its hard to wrap my head around whats going on.
So I've got to the point where I have expanded (ax + b)^n, I have differentiated it and I have (a) as a factor of (ax + b)^(n-1), except with (n) as a factor of the first term (ax)^(n-1). Now there's no way I can post all of my work, which includes Pascal notation or whatever the conventional term is.
Anyway I'd be thankful if someone having the appropriate software could provide me with a proof so I can see where I've gone wrong.
Thanks

I don't agree with JeffM that you are required to use the limit process for this. [Nothing wrong with his answer, though.] But just using the x^n rule and binomial expansion, I think it goes like this:

In the expansion:

(ax + b)^n = SUM[k=0 to n] C(n,k)(ax)^(n-k) b^k

[Note that the k=n term is a constant.]

Now differentiate each term:

dy/dx = SUM[k=0 to n] (n-k) C(n,k)(ax)^(n-k-1)(a) b^k

Now note that the k=n term is zero, so that is:

dy/dx = SUM[k=0 to n-1] (n-k) C(n,k)(ax)^(n-k-1)(a) b^k

.....................

Now

C(n,k) = n!/[(n-k)! k!]

and

(n-k)C(n,k) = n!(n-k)/[(n-k)! k!]

= n!/[ (n-k-1)! k! ]

= n (n-1)!/ [ (n-k-1)! k! ]

= n (n-1)! / [ (n-1-k)! k! ]

= n C(n-1,k)

SO:

dy/dx = SUM[k=0 to n-1] (n-k) C(n,k)(ax)^(n-k-1)(a) b^k

becomes

dy/dx = SUM[k=0 to n-1] n C(n-1,k)(ax)^(n-1-k)(a) b^k

factoring out na:

dy/dx = an SUM[k=0 to n-1] C(n-1,k)(ax)^(n-1-k) b^k

and I think the SUM is the expansion of

dy/dx = an (ax + b)^n-1

 

[SAMUELK]

Edit: No, the simplest proof is to use induction, the chain rule, and the product rule. I do not understand problems that ask you to use a screw driver to hammer nails.

You are correct, of course -- this is the simplest way. But the teacher said to use the expansion, and, to paraphrase an old musical,

Whatever teacher wants, teacher gets.